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Problem 1863 Sum of All Subset XOR Totals

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Backtracking Bit-Manipulation Recursion Cpp

Problem Statement

Link - Problem 1863

Question

The XOR total of an array is defined as the bitwise XOR of all its elements, or 0 if the array is empty.

For example, the XOR total of the array [2,5,6] is 2 XOR 5 XOR 6 = 1. Given an array nums, return the sum of all XOR totals for every subset of nums.

Note: Subsets with the same elements should be counted multiple times.

An array a is a subset of an array b if a can be obtained from b by deleting some (possibly zero) elements of b.

Example 1

Input: nums = [1,3]
Output: 6
Explanation: The 4 subsets of [1,3] are:
- The empty subset has an XOR total of 0.
- [1] has an XOR total of 1.
- [3] has an XOR total of 3.
- [1,3] has an XOR total of 1 XOR 3 = 2.
0 + 1 + 3 + 2 = 6

Example 2

Input: nums = [5,1,6]
Output: 28
Explanation: The 8 subsets of [5,1,6] are:
- The empty subset has an XOR total of 0.
- [5] has an XOR total of 5.
- [1] has an XOR total of 1.
- [6] has an XOR total of 6.
- [5,1] has an XOR total of 5 XOR 1 = 4.
- [5,6] has an XOR total of 5 XOR 6 = 3.
- [1,6] has an XOR total of 1 XOR 6 = 7.
- [5,1,6] has an XOR total of 5 XOR 1 XOR 6 = 2.
0 + 5 + 1 + 6 + 4 + 3 + 7 + 2 = 28

Example 3

Input: nums = [3,4,5,6,7,8]
Output: 480
Explanation: The sum of all XOR totals for every subset is 480.

Constraints

Solution

class Solution {
public:
    int subsetXORSum(vector<int>& nums) {
        std::ios::sync_with_stdio(false);
        return calculate(nums, 0, 0);
    }

    int calculate(vector<int>& nums, int level, int currentXOR) {

        if (level == nums.size())
            return currentXOR;

        int include = calculate(nums, level + 1, currentXOR ^ nums[level]);

        int exclude = calculate(nums, level + 1, currentXOR);

        return include + exclude;
    }
};

Complexity Analysis

Explanation

1. Intuition

2. Implementation

3. Functions


This problem demonstrates the use of recursion, backtracking and bit manipulation to solve combinatorial problems.